# Thread: Sine/ Cosine Rule Problems

1. ## Sine/ Cosine Rule Problems

Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $\displaystyle x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.

Thanks!

2. Originally Posted by ns2583

Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $\displaystyle x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.

Thanks!
Your equation is correct.

You can solve it for $\displaystyle x$ using the Quadratic formula.

3. Originally Posted by ns2583

Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $\displaystyle x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.

Thanks!
(a) is a direct application of the cosine rule: $\displaystyle 10^2 = 7^2 + x^2 - 2 (x)(7) \cos 50^0$.

(b) $\displaystyle \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\displaystyle \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\displaystyle \frac{7}{\sin \beta} = \frac{10}{\sin 50^0}$ .... (2)

From (2) get $\displaystyle \beta$. From $\displaystyle \alpha + \beta + 50^0 = 180^0$ get $\displaystyle \alpha$. Now use (1) to solve for $\displaystyle x$.

4. Originally Posted by Prove It
Your equation is correct.

You can solve it for $\displaystyle x$ using the Quadratic formula.
When I put it in my calc, it said there was a nonreal answer. :S

Originally Posted by mr fantastic
(a) is a direct application of the cosine rule:

(b) $\displaystyle \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\displaystyle \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\displaystyle \frac{7}{\sin \beta} = \frac{10}{\sin 50^0}$ .... (2)

From (2) get $\displaystyle \beta$. From $\displaystyle \alpha + \beta + 50^0 = 180^0$ get $\displaystyle \alpha$. Now use (1) to solve for $\displaystyle x$.
Got it, thanks!

5. Originally Posted by ns2583
Originally Posted by Prove It
Your equation is correct.

You can solve it for $\displaystyle x$ using the Quadratic formula.
When I put it in my calc, it said there was a nonreal answer. :S
Huh? Just use the quadratic formula:

\displaystyle \begin{aligned} x^2 - 14(\cos 50^{\circ}) x - 51 &= 0 \\ x^2 - 9x - 51 &\approx 0 \\ x &\approx \frac{9 \pm \sqrt{(-9)^2 - 4(1)(-51)}}{2(1)} \\ &\approx \frac{9 \pm \sqrt{284.98}}{2} \\ \end{aligned}

$\displaystyle x \approx 12.94$
$\displaystyle x \approx -3.94$
Reject the 2nd solution, so x is about 13 cm.

01

6. OH, i see. Thanks.